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White raven [17]
3 years ago
6

When you drain the solution from the bottom, the concentration will:

Chemistry
1 answer:
djverab [1.8K]3 years ago
3 0

.Answer:

stay the same the conc. of a solution is the same throughout the liquid

Explanation:

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During science, Jim's teacher gives this simple rule to use as a first attempt to increase solubility of the solute and the rate
gladu [14]
B. Agitate the solute in the solvent.
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3 years ago
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At 350 k, kc = 0.142 for the reaction 2 brcl(g) ⇀↽ br2(g) + cl2(g) an equilibrium mixture at this temperature contains equal con
djyliett [7]
0.114 mol/l  
The equilibrium equation will be: 
Kc = ([Br2][Cl2])/[BrCl]^2  
The square factor for BrCl is due to the 2 coefficient on that side of the equation.  
Now solve for BrCl, substitute the known values and calculate. 
Kc = ([Br2][Cl2])/[BrCl]^2 
[BrCl]^2 * Kc = ([Br2][Cl2]) 
[BrCl]^2 = ([Br2][Cl2])/Kc 
[BrCl] = sqrt(([Br2][Cl2])/Kc)  
[BrCl] = sqrt(0.043 mol/l * 0.043 mol/l / 0.142) 
[BrCl] = sqrt(0.001849 mol^2/l^2 / 0.142) 
[BrCl] = sqrt(0.013021127 mol^2/l^2) 
[BrCl] = 0.114110152 mol/l  
Rounding to 3 significant figures gives 0.114 mol/l
4 0
3 years ago
Which of the following organisms would NOT be in the first trophic level of an energy pyramid?
faust18 [17]
dog!

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7 0
3 years ago
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Give an example of a solution that can be seperated by simple distilation.​
Juliette [100K]

Answer: Evaporation of salt water

Explanation: During the process of evaporating salt water which involves simple distillation, pure water is separated with salt molecules.

When heated water evaporates from the solution since it is less dense. When condensed it becomes pure water and salt is left out since it is more denser.

8 0
3 years ago
Q Q 3. (08.02 MC)
AnnyKZ [126]

Answer: A volume of 455 mL from 0.550 M KBr solution can be made from 100.0 mL of 2.50 M KBr.

Explanation:

Given: V_{1} = ?,         M_{1} = 0.55 M

V_{2} = 100.0 mL,        M_{2} = 2.50 M

Formula used to calculate the volume of KBr is as follows.

M_{1}V_{1} = M_{2}V_{2}

Substitute the values into above formula as follows.

M_{1}V_{1} = M_{2}V_{2}\\0.55 M \times V_{1} = 2.50 M \times 100.0 mL\\V_{1} = 455 mL

Thus, we can conclude that a volume of 455 mL from 0.550 M KBr solution can be made from 100.0 mL of 2.50 M KBr.

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3 years ago
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